Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector .
The vector form of the equation of plane is expressed as,
r → ⋅ n ^ = d (1)
Here,
n ^ is the normal unit vector of the plane.
d is the distance of the plane from origin.
The normal vector is expressed as,
n → = 3 i ^ + 5 j ^ − 6 k ^
The magnitude of the normal vector is,
| n → | = 3 2 + 5 2 + ( − 6 ) 2 = 9 + 25 + 36 = 70
The normal unit vector is,
n ^ = n → | n → | = 3 i ^ + 5 j ^ − 6 k ^ 70 (2)
The distance of the plane from origin is,
d = 7 units (3)
Substitute equation (2) and (3) in equation (1).
r → ⋅ ( 3 i ^ + 5 j ^ − 6 k ^ 70 ) = 7 r → ⋅ ( 3 i ^ + 5 j ^ − 6 k ^ ) = 7 70
Thus, the vector equation of the plane is r → ⋅ ( 3 i ^ + 5 j ^ − 6 k ^ ) = 7 70 .
The vector form of the equation of plane is expressed as,
Here,
The normal vector is expressed as,
The magnitude of the normal vector is,
The normal unit vector is,
The distance of the plane from origin is,
Substitute equation (2) and (3) in equation (1).
Thus, the vector equation of the plane is
.